Table of Contents

Toggle**NCERT Solutions for class 6, Maths chapter 5** has basic learning concepts of various shapes. This class 6 maths exercise 5.1 is about knowledge of line segments and basic requirements of making triangle.The questions in this exercise named as Understanding Elementary Shapes, Exercise 5.1 given below.

## Understanding Elementary Shapes, Exercise 5.1

**Q.1. What is the disadvantage in comparing line segments by mere observation?**

**Ans: **It causes occurrence of errors because by mere observation , we can’t compare the line segments with slight difference in their length. We can’t say which line segment is of greater length. Hence, the chances of errors are more.

**Q.2. Why is it better to use a divider than a ruler, while measuring the length of a line segment?**

**Ans: **Measurement by a ruler may be not accurate because of the thickness of the ruler. Hence, it is better to use a divider than a ruler, while measuring the length of a line segment.

**Q.3. Draw any line segment, say ** **. Take any point C lying in between A and B. Measure the lengths of AB, BC and AC. Is AB = AC + CB ? [Note : If A,B,C are any three points on a line such that AC + CB = AB, then we can be sure that C lies between A and B.]**

**Ans: **Yes, the length of line segment AB is equal to sum of line segment AC and CB.

Therefore, for every situation in which point C is lying in between A and B we may say that

AB = AC + CB

For example:

AB = 8 cm and C is a point between A and B such that AC = 5 cm and CB = 3 cm.

Hence, AC + CB = (5+3) cm = 8 cm

Since, AB = 8 cm

∴ **AB = AC + CB is verified.**

**Q.4.If A,B,C ****are three points on a line such that AB = 5 cm, BC = 3 cm and AC = 8 cm, which one of them lies between the other two?**

**Ans: **Given, AB = 5 cm**, **BC = 3 cm and AC = 8 cm

is the longest line segment, thus B is the point between A and C.

Therefore, AC = AB + BC

Thus, point B lies between A and C.

**Q.5. Verify, whether D is the mid point of $\overline {AG} $** ** .**

**Ans: **Here, AD = 3 units and DG = 3 units

∴ AD = DG

Hence, D is the midpoint of** .**

**Q.6. If B is the mid point of AC and C is the mid point of ** ** , where A,B,C,D lie on a straight line, say why AB = CD?**

**Ans:**

Given

B is the midpoint of . Hence, AB = BC ……………………. (i)

And C is the midpoint of . Hence, BC = CD ……………….(ii)

From (1) and (2), we get

AB = CD (Hence Verified)

**Q.7. Draw five triangles and measure their sides. Check in each case, if the sum of the lengths of any two sides is always less than the third side.**

**Ans: ****(a) In triangle ABC**

AB= 3 cm , BC = 4 cm and AC = 5 cm

AB + BC = (3 + 4) cm = 7 cm

As, 7 > 5

∴ AB + BC > AC

Hence, the sum of any two sides of a triangle is greater than the third side.

**(b) In triangle PQR**

PQ = 8 cm, QR = 5 cm and PR = 10 cm

PQ + QR = (8 + 5) cm = 13 cm

As 13 > 10

∴ PQ + QR > PR

Hence, the sum of any two sides of a triangle is greater than the third side.

**(c) In triangle KLM**

KL = 8 cm, LM = 8 cm and KM = 8 cm

KL + LM = (8+8) cm = 16 cm

As 16 cm > 8 cm

∴ KL + LM > KM

Hence, the sum of any two sides of a triangle is greater than the third side.

**(d) In triangle ABC:**

AB= 3.5 cm , BC = 5.5 cm and AC = 8.5 cm

AB + BC = (3.5 + 5.5) cm = 9 cm

As, 9 > 8.5

∴ AB + BC > AC

Hence, the sum of any two sides of a triangle is greater than the third side.**(e) In triangle PQR**

PQ = 2.5 cm, QR = 3.5 cm and PR = 4.5 cm

PQ + QR = (2.5 + 3.5) cm = 6 cm

As 6 > 4.5

∴ PQ + QR > PR

Hence, the sum of any two sides of a triangle is greater than the third side.

Therefore, we conclude that the sum of any two sides of a triangle is always greater than the third side.

## Chapter 5 : Understanding Elementary Shapes

**Understanding Elementary Shapes, Exercise 5.1**

**Understanding Elementary Shapes, Exercise 5.2**

**Understanding Elementary Shapes, Exercise 5.3**

**Understanding Elementary Shapes, Exercise 5.4**

**Understanding Elementary Shapes, Exercise 5.5**

**Understanding Elementary Shapes, Exercise 5.6**

**Understanding Elementary Shapes, Exercise 5.7**